题解 | #牛牛的新数组求和#

#include <stdio.h>
#include <stdlib.h>


int cal(int* arry,int n)
{
        int sum = 0;
        while(* arry != '\0')
        {
                sum += * arry;
                arry++;
        }

        return sum;

}
int main() 
{
    int arry[100] ={0};
    int n;
    int sum;
    while(scanf("%d",&n) != EOF)
    {
         for(int i =0;i<n;i++)
         {
             scanf("%d",&arry[i]);
         }

        sum=cal(arry,n);

        printf("%d",sum);
    }

    return 0;
}

rry为数组的首地址

用sum接受cal的返回值并输出